A circular disc of radius R and thickness fra | vedclass

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(C) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.Since

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$\frac{I}{5}$

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(C) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.Since the disc is melted and recasted into a solid sphere, the volume remains constant.Volume of disc = Volume of sphere$\pi R^2 	imes \frac{R}{6} = \frac{4}{3} \pi R_1^3$$\frac{R^3}{6} = \frac{4}{3} R_1^3$$R_1^3 = \frac{R^3}{8} \implies R_1 = \frac{R}{2}$, where $R_1$ is the radius of the sphere.The moment of inertia of a solid sphere about its diameter is $I' = \frac{2}{5} M R_1^2$.Substituting $R_1 = \frac{R}{2}$ into the formula:$I' = \frac{2}{5} M \left(\frac{R}{2}ight)^2 = \frac{2}{5} M \left(\frac{R^2}{4}ight) = \frac{1}{5} \left(\frac{1}{2} M R^2ight)$.Since $I = \frac{1}{2} M R^2$, we get $I' = \frac{I}{5}$.

$A$ circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is :

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